2Noetherian algebras
III Algebras
2.2 More on A
n
(k) and U(g)
Our goal now is to study modules of
A
n
(
k
). The first result tells us we must
focus on infinite dimensional modules.
Lemma.
Suppose
char k
= 0. Then
A
n
(
k
) has no non-zero modules that are
finite-dimensional k-vector spaces.
Proof.
Suppose
M
is a finite-dimensional module. Then we’ve got an algebra
homomorphism θ : A
n
(k) → End
k
(M)
∼
=
M
m
(k), where m = dim
k
M.
In A
n
(k), we have
Y
1
X
1
− X
1
Y
1
= 1.
Applying the trace map, we know
tr(θ(Y
1
)θ(X
1
) − θ(X
1
)θ(Y
1
)) = tr I = m.
But since the trace is cyclic, the left hand side vanishes. So
m
= 0. So
M
is
trivial.
A similar argument works for the quantum torus, but using determinants
instead.
We’re going to make use of our associated graded algebras from last time,
which are isomorphic to polynomial algebras. Given a filtration
{A
i
}
of
A
, we
may filter a module with generating set S by setting
M
i
= A
i
S.
Note that
A
j
M
i
⊆ M
i+j
,
which allows us to form an associated graded module
gr M =
M
M
i
M
i+1
.
This is a graded
gr A
-module, which is finitely-generated if
M
is. So we’ve got a
finitely-generated graded module over a graded commutative algebra.
To understand this further, we prove some results about graded modules
over commutative algebras, which is going to apply to our gr A and gr M.
Definition
(Poincar´e series)
.
Let
V
be a graded module over a graded algebra
S, say
V =
∞
M
i=0
V
i
.
Then the Poincar´e series is
P (V, t) =
∞
X
i=0
(dim V
i
)t
i
.
Theorem
(Hilbert-Serre theorem)
.
The Poincar´e series
P
(
V, t
) of a finitely-
generated graded module
V =
∞
M
i=0
V
i
over a finitely-generated generated commutative algebra
S =
∞
M
i=0
S
i
with homogeneous generating set x
1
, · · · , x
m
is a rational function of the form
f(t)
Q
(1 − t
k
i
)
,
where f(t) ∈ Z[t] and k
i
is the degree of the generator x
i
.
Proof.
We induct on the number
m
of generators. If
m
= 0, then
S
=
S
0
=
k
,
and
V
is therefore a finite-dimensional
k
-vector space. So
P
(
V, t
) is a polynomial.
Now suppose
m >
0. We assume the theorem is true for
< m
generators.
Consider multiplication by x
m
. This gives a map
V
i
V
i+k
m
x
m
,
and we have an exact sequence
0 K
i
V
i
V
i+k
m
L
i+k
m
→ 0,
x
m
(∗)
where
K =
M
K
i
= ker(x
m
: V → V )
and
L =
M
L
i+k
m
= coker(x
m
: V → V ).
Then
K
is a graded submodule of
V
and hence is a finitely-generated
S
-module,
using the fact that
S
is Noetherian. Also,
L
=
V/x
m
V
is a quotient of
V
, and it
is thus also finitely-generated.
Now both
K
and
L
are annihilated by
x
m
. So they may be regarded as
S
0
[x
1
, · · · , x
m−1
]-modules. Applying dim
k
to (∗), we know
dim
k
(K
i
) − dim
k
(V
i
) + dim(V
i+k
m
) − dim(L
i+k
m
) = 0.
We multiply by t
i+k
m
, and sum over i to get
t
k
m
P (K, t) − t
k
m
P (V, t) + P (V, t) − P (L, t) = g(t),
where
g
(
t
) is a polynomial with integral coefficients arising from consideration
of the first few terms.
We now apply the induction hypothesis to K and L, and we are done.
Corollary.
If each
k
1
, · · · , k
m
= 1, i.e.
S
is generated by
S
0
=
k
and homoge-
neous elements
x
1
, · · · , x
m
of degree 1, then for large enough
i
, then
dim V
i
=
φ
(
i
)
for some polynomial
φ
(
t
)
∈ Q
[
t
] of
d −
1, where
d
is the order of the pole of
P (V, t) at t = 1. Moreover,
i
X
j=0
dim V
j
= χ(i),
where χ(t) ∈ Q[t] of degree d.
Proof. From the theorem, we know that
P (V, t) =
f(t)
(1 − t)
d
,
for some d with f (1) 6= 0, f ∈ Z[t]. But
(1 − t)
−1
= 1 + t + t
2
+ · · ·
By differentiating, we get an expression
(1 − t)
−d
=
X
d + i − 1
d − 1
t
i
.
If
f(t) = a
0
+ a
1
t + · · · + a
s
t
s
,
then we get
dim V
i
= a
0
d + i − 1
d − 1
+ a
1
d + i − 2
d − 1
+ · · · + a
s
d + i − s − 1
d − 1
,
where we set
r
d−1
= 0 if
r < d −
1, and this expression can be rearranged to
give φ(i) for a polynomial φ(t) ∈ Q[t], valid for i − s > 0. In fact, we have
φ(t) =
f(1)
(d − 1)!
t
d−1
+ lower degree term.
Since f(1) 6= 0, this has degree d − 1.
This implies that
i
X
j=0
dim V
j
is a polynomial in Q[t] of degree d.
This
φ
(
t
) is the Hilbert polynomial, and
χ
(
t
) the Samuel polynomial . Some
people call χ(t) the Hilbert polynomial instead, though.
We now want to apply this to our cases of
gr A
, where
A
=
A
n
(
k
) or
U
(
g
),
filtered as before. Then we deduce that
i
X
0
dim
M
j
M
j−1
= χ(i),
for a polynomial χ(t) ∈ Q[t]. But we also know
i
X
j=0
dim
M
j
M
j−1
= dim M
i
.
We are now in a position to make a definition.
Definition
(Gelfand-Kirillov dimension)
.
Let
A
=
A
n
(
k
) or
U
(
g
) and
M
a
finitely-generated
A
-module, filtered as before. Then the Gelfand-Kirillov di-
mension
d
(
M
) of
M
is the degree of the Samuel polynomial of
gr M
as a
gr A-module.
This makes sense because
gr A
is a commutative algebra in this case. A
priori, it seems like this depends on our choice of filtering on
M
, but actually, it
doesn’t. For a more general algebra, we can define the dimension as below:
Definition
(Gelfand-Kirillov dimension)
.
Let
A
be a finitely-generated
k
-
algebra, which is filtered as before, and a finitely-generated
A
-module
M
, filtered
as before. Then the GK-dimension of M is
d(M) = lim sup
n→∞
log(dim M
n
)
log n
.
In the case of
A
=
A
n
(
k
) or
U
(
g
), this matches the previous definition. Again,
this does not actually depend on the choice of generating sets.
Recall we showed that no non-zero
A
n
(
k
)-module
M
can have finite dimension
as a
k
-vector space. So we know
d
(
M
)
>
0. Also, we know that
d
(
M
) is an
integer for cases
A
=
A
n
or
U
(
g
), since it is the degree of a polynomial. However,
for general
M
=
A
, we can get non-integral values. In fact, the values we can
get are 0
,
1
,
2, and then any real number
≥
2. We can also have
∞
if the
lim sup
doesn’t exist.
Example.
If
A
=
kG
, then we have
GK
-
dim
(
kG
)
< ∞
iff
G
has a subgroup
H
of finite index with
H
embedding into the strictly upper triangular integral
matrices, i.e. matrices of the form
1 ∗ · · · ∗
0 1 · · · ∗
.
.
.
.
.
.
.
.
.
.
.
.
0 0 · · · 1
.
This is a theorem of Gromoll, and is quite hard to prove.
Example.
We have
GK
-
dim
(
A
) = 0 iff
A
is finite-dimensional as a
k
-vector
space.
We have
GK-dim(k[X]) = 1,
and in general
GK-dim(k[X
1
, · · · , X
n
]) = n.
Indeed, we have
dim
k
(mth homogeneous component) =
m + n
n
.
So we have
χ(t) =
t + n
n
This is of degree n, with leading coefficient
1
n!
.
We can make the following definition, which we will not use again:
Definition
(Multiplicity)
.
Let
A
be a commutative algebra, and
M
an
A
-module.
The multiplicity of M with d(M) = d is
d! × leading coefficient of χ(t).
On the second example sheet, we will see that the multiplicity is integral.
We continue looking at more examples.
Example.
We have
d
(
A
n
(
k
)) = 2
n
, and
d
(
U
(
g
)) =
dim
k
g
. Here we are using
the fact that the associated graded algebras are polynomial algebras.
Example.
We met
k
[
X
1
, · · · , X
n
] as the “canonical”
A
n
(
k
)-module. The filtra-
tion of the module matches the one we used when thinking about the polynomial
algebra as a module over itself. So we get
d(k[X
1
, · · · , X
n
]) = n.
Lemma. Let M be a finitely-generated A
n
-module. Then d(M) ≤ 2n.
Proof.
Take generators
m
1
, · · · , m
s
of
M
. Then there is a surjective filtered
module homomorphism
A
n
⊕ · · · ⊕ A
n
M
(a
1
, · · · , a
s
)
P
a
i
m
i
It is easy to see that quotients can only reduce dimension, so
GK-dim(M) ≤ d(A
n
⊕ · · · ⊕ A
n
).
But
χ
A
n
⊕···⊕A
n
= sχ
A
n
has degree 2n.
More interestingly, we have the following result:
Theorem
(Bernstein’s inequality)
.
Let
M
be a non-zero finitely-generated
A
n
(k)-module, and char k = 0. Then
d(M) ≥ n.
Definition
(Holonomic module)
.
An
A
n
(
k
) module
M
is holonomic iff
d
(
M
) =
n.
If we have a holonomic module, then we can quotient by a maximal submodule,
and get a simple holonomic module. For a long time, people thought all simple
modules are holonomic, until someone discovered a simple module that is not
holonomic. In fact, most simple modules are not holonomic, but we something
managed to believe otherwise.
Proof.
Take a generating set and form the canonical filtrations
{A
i
}
of
A
n
(
k
)
and
{M
i
}
of
M
. We let
χ
(
t
) be the Samuel polynomial. Then for large enough
i, we have
χ(i) = dim M
i
.
We claim that
dim A
i
≤ dim Hom
k
(M
i
, M
2i
) = dim M
i
× dim M
2i
.
Assuming this, for large enough i, we have
dim A
i
≤ χ(i)χ(2i).
But we know
dim A
i
=
i + 2
2n
,
which is a polynomial of degree 2
n
. But
χ
(
t
)
χ
(2
t
) is a polynomial of degree
2d(M). So we get that
n ≤ d(M ).
So it remains to prove the claim. It suffices to prove that the natural map
A
i
→ Hom
k
(M
i
, M
2i
),
given by multiplication is injective.
So we want to show that if
a ∈ A
i
6
= 0, then
aM
i
6
= 0. We prove this by
induction on
i
. When
i
= 0, then
A
0
=
k
, and
M
0
is a finite-dimensional
k
-vector
space. Then the result is obvious.
If
i >
0, we suppose the result is true for smaller
i
. We let
a ∈ A
i
is non-zero.
If aM
i
= 0, then certainly a 6∈ k. We express
a =
X
c
αβ
X
α
1
1
X
α
2
2
· · · X
α
n
n
Y
β
1
1
· · · Y
β
n
n
,
where α = (α
1
, · · · , α
n
), β = (β
1
, · · · , β
n
), and c
α,β
∈ k.
If possible, pick a
j
such that
c
α,α
6
= 0 for some
α
with
α
j
6
= 0 (this happens
when there is an X involved). Then
[Y
j
, a] =
X
α
j
c
α,β
X
α
1
1
· · · X
α
j
−1
j
· · · X
α
n
n
Y
β
1
1
· · · Y
β
n
n
,
and this is non-zero, and lives in A
i−1
.
If aM
i
= 0, then certainly aM
i−1
= 0. Hence
[Y
j
, a]M
i−1
= (Y
j
a − aY
j
)M
i−1
= 0,
using the fact that Y
j
M
i−1
⊆ M
i
. This is a contradiction.
If a only has Y ’s involved, then we do something similar using [X
j
, a].
There is also a geometric way of doing this.
We take k = C. We know gr A
n
is a polynomial algebra
gr A
n
= k[
¯
X
1
, · · · ,
¯
X
n
,
¯
Y
1
, · · · ,
¯
Y
n
],
which may be viewed as the coordinate algebra of the cotangent bundle on affine
n
-space
C
n
. The points of this correspond to the maximal ideals of
gr A
n
. If
I
is a left ideal of
A
n
(
C
), then we can form
gr I
and we can consider the set of
maximal ideals containing it. This gives us the characteristic variety
Ch
(
A
n
/I
).
We saw that there was a Poisson bracket on
gr A
n
, and this may be used to
define a skew-symmetric form on the tangent space at any point of the cotangent
bundle. In this case, this is non-degenerate skew-symmetric form.
We can consider the tangent space
U
of
Ch
(
A
n
/I
) at a non-singular point,
and there’s a theorem of Gabber (1981) which says that
U ⊇ U
⊥
, where
⊥
is
with respect to the skew-symmetric form. By non-degeneracy, we must have
dim U ≥ n, and we also know that
dim Ch(A
n
/I) = d(A
n
/I).
So we find that d(A
n
/I) ≥ n.
In the case of
A
=
U
(
g
), we can think of
gr A
as the coordinate algebra on
g
∗
, the vector space dual of
g
. The Poisson bracket leads to a skew-symmetric
form on tangent spaces at points of
g
∗
. In this case, we don’t necessarily get
non-degeneracy. However, on
g
, we have the adjoint action of the corresponding
Lie group
G
, and this induces a co-adjoint action on
g
∗
. Thus
g
∗
is a disjoint
union of orbits. If we consider the induced skew-symmetric form on tangent
spaces of orbits (at non-singular points), then it is non-degenerate.